实验22单源最短路径

题目：p3371​​​​​​

解题思路：dijkstra算法+邻接表

代码：

#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
const int N = 10001;
const int inf = 0x3f3f3f3f;
typedef struct edge {
	int to;//一条边的终点
	int w;//边权重
}Edge;
vector<Edge> G[N];//数组的每一个元素相当于一个vector
int dist[N];//记录所有点到源点的最短距离
int visited[N];//记录是否访问过这一点
int n, m, s;//n：点的个数 m：边的个数 s起点
void dij();
int main()
{
	cin >> n >> m >> s;
	Edge temp1, temp2;
	int x, y, z;//起点 终点 权重
	memset(visited, 0, sizeof visited);
	memset(dist, 0x3f, sizeof dist);
	for (int i = 1; i <= m; i++)
	{
		cin >> x >> y >> z;
		temp1.to = y; temp1.w = z;
		G[x].push_back(temp1);
	}
	 
	dist[s] = 0;//源点到源点的距离最小值一定是0
	dij();
	for (int i = 1; i <= n; i++)
	{
		if (dist[i] == inf)//inf表示两个点不连通，根据题意输出INT_MAX
		{
			cout << 2147483647<<" ";
		}
		else
			cout << dist[i] << " ";
	}
}

void dij()
{
	int minn, index;
	while (1)//当所有的点都遍历了一遍后，跳出循环
	{
		minn = inf; index = -1;
		for (int i = 1; i <= n; i++)
		{
			if (!visited[i] && dist[i] < minn)
			{
				minn = dist[i];
				index = i;
			}
		}
		if (index == -1)break;//如果遍历了所有点后都没有改变index的值
		visited[index] = 1;
		int len = G[index].size();
		int end, w;
		for (int i = 0; i < len; i++)
		{
			end = G[index][i].to;
			w = G[index][i].w;
			if (!visited[end] && dist[end] > w + dist[index])
			{
				dist[end] = w + dist[index];
			}
		}
	}
}